Question

How do you find the solutions of cos^1x - 3sinx + 2sin^2x = 0 in degrees?http://bamboodock.wacom.com/doodler/6f06a9ee-6b26-435d-9685-2e62bcaa13b5
http://bamboodock.wacom.com/doodler/5ebb5618-f075-482f-88ac-3645f2ed822d

Answer 1

i mean cos^2x - 3sinx + 2sin^2x = 0

Answer 2

i think u did it wrong

Answer 3

This will become:

\[\sin^2(x) - 3\sin(x) + 1 = 0\]

Answer 4

sin(x) (2 sin(x)-3)+cos(x) = 0

Answer 5

-3 sin(x)+cos(x)-cos(2 x) = -1

Answer 6

@waterineyes is doing right

Answer 7

-3 sin(x)+cos(x)-cos(2 x)+1 = 0

Answer 8

now u can do it easily

Answer 9

ans
2 sin^2(x)-3 sin(x)+cos(x) = 0

Answer 10

\[\sin(x) = \frac{3 \pm \sqrt{5}}{2}\]

Answer 11

wait so my first step is to convert cos^2 to sin right?

Answer 12

Yes using:

\[\cos^2(x) = 1 - \sin^2(x)\]